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3 years ago

Consider a horse pulling a buggy. Is the

Physics

1 answer:

Degger [83]3 years ago

8 0

Yes, it's true.

But 2nd Newton Law always come to play when the horse is to move forward because obviously the forces interact antagonistically and mass has to be accounted for.

That's what I think. Hope it's right, all the best.

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What are the equivalent celsius and kelvin temperatures of -128 f?

Gre4nikov [31]

Answer:

128 Kelvin = 128 - 273.15 = -145.15 Celsius. Temperature conversion chart Sample temperature conversions 103.55 Kelvin to degrees Fahrenheit 39.82 degrees Fahrenheit to Kelvin

Explanation:

hope this helps have a good day

6 0

2 years ago

light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

lapo4ka [179]

Answer:

The distance traveled in 1 year is: $3.143*10^{16}ft$

Explanation:

Given

$s = 982,080,000 ft/s$ --- speed

$t = 32,000,000 s$ --- time

Required

The distance traveled

This is calculated as:

$Speed = \frac{Distance}{Time}$

So, we have:

$Distance = Speed * Time$

This gives:

$Distance = 982,080,000 ft/s * 32,000,000 s$

$Distance = 982,080,000 * 32,000,000ft$

$Distance = 3.143*10^{16}ft$ -- approximated

5 0

3 years ago

What organs make up the skeletal system

USPshnik [31]

<span>Bones. The most important organ of the skeletal system is the bones.

Ligaments and Joints. Another important component, i.e. the ligaments are made of fibrous collagen tissue that attaches one bone to another bone.
<span>
Cartilage.</span></span>

5 0

3 years ago

Read 2 more answers

Calculate the wavelength of an orange light wave with a frequency of 5.085 x 10^14 Hz. The speed of light is 3.0 x 10^8 m/s.

Margarita [4]

Answer:

5.9 x 10⁻⁷m

Explanation:

Given parameters:

Frequency = 5.085 x 10¹⁴Hz

Speed of light = 3.0 x 10⁸m/s

Unknown:

Wavelength of the orange light = ?

Solution:

The wavelength can be derived using the expression below;

wavelength = $\frac{v}{f}$

v is the speed of light

f is the frequency

wavelength = $\frac{3 x 10^{8} }{5.085 x 10^{14} }$ = 5.9 x 10⁻⁷m

3 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago